"Emma has 5 French books, 6 English books and 4 Spanish books on her shelf. Find the probability that if she chooses four books at random, she will have at least one book of each language."

Seriously, if you can do this, I will love you forever. My brother- a 3rd Year engineering student who just did a course on probability, couldn't get it- but he has a tendency to overcomplicate. The answer should be 48/91- but the marking scheme just randomly multiplies everything for 4!/2! to get that and I have no idea why.

-Stella x

Absolutely not a ninety year old German lady masquerading as an Irish med student.

"Stella. You were in my dream the other night. And everyone called you Princess." -Lauren2010

I can do this! I think. So, you're drawing four, right? Assuming you do not replace any of the books, it should look something like this:

You have 15 books total, and you want one of each.

So, to start with. Let's say you want to do the French book first. So your odds on the first draw are 5/15. If you do English next, you have 6/14 - 6 English books total, out of 14 books left - assuming you've taken one French. Then you'll have 4/13 for Spanish books. Which is 12/84. Nvm.

Um. OH! Fourth draw. Ok. Then we have (4+5+3)/12. Lemme see if that works out - no.

NO. Ok. I think what i was doing first is right, I just didn't take it far enough. Ok.

So for French, you must have at least one, so 5/15. But your second draw could be French, too, so that'd be 4/14. But you can't do more, since you must have at least one in the other languages, too. So next we have English - 6/15. Then 5/14. Then finally Spanish - 4/15, and then 3/14. Erm.

Don't ask me to simplify the mess of numbers for you, however. So, the bit in brackets is your numerator. You multiply all the bits together, then add them up - so like the first C(5,2) C(6,1) C(4,1) bit here, you multiply it together once you've found the combinations. I just divided my final numbers, and got 720/13365 = .527. Same as 48/91 = .527 (same through at least 10 decimal points, so good enough for me. I'm just lazy and didn't put them all down).

So, the logic here!

You have four picks, and must have at least one of three different types of books. So, one type of book MUST have two, as you have 4 picks and 3 types. So, each grouping accounts for this - your first event has you picking two French books, and then oen of the other two. Combination, as order doesn't matter like it would in a permutation, you're not replacing books, and you draw together - I don't quite remember the logic behind this end, but I'll look it up in my notes if you need me to! It has to do with double-counting somewhere. Anyway, you do that for each language (so two English and two Spanish respectively). You then put all of that over your total possible picks - C(15,4) - you have 15 books to choose from, and want 4 of them. Since order doesn't matter, combination still - any four books will do here, as it were, so you're possible outcomes are all possible combinations of four books.

Anyway, I hope that made some sense! Just ignore all my silliness if figuring it out, since I had to reason my way through - it's been a few months since I've looked at probability problems!

***Under the Responsibility of S.P.E.W.***(Sadistic Perplexion of Everyone's Wits)

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